The activation energy of a reaction is zero. Its rate constant at 280 K is `16 xx 10^(-6) s^(-1)` the rate constant at 300k is

Options

A. `3.2 xx 10^(-6) s^(-1)`
B. zero
C. `1.6 xx 10^(-6)s^(-1)`
D. `1.6 xx 10^(-5)s^(-1)`

Answer

D

Explanation

`"log" (k_(2))/(k_(1)) = (E_(a))/(2.303 R) [(1)/(T_(1)) - (1)/(T_(2))]`
Given,
Activation energy of a reaction, `E_(A) = 0`
Rate constant, `k_(1) = 1.6 xx 10^(-6) s^(-1)`
Temperature, `T_(1) = 280 K, T_(2) = 300 K`
According to Arrhenius equation
`"log" (k_(2))/(1.6 xx 10^(-6)) = (0)/(2.303 R) [(1)/(280) - (1)/(300)]`
`"log" (k_(2))/(1.6 xx 10^(-6)) = 0`
`(k_(2))/(1.6 xx 10^(-6))` = antilog 0
`(k_(2))/(1.6 xx 10^(-6)) = 1`
`:. k_(2) = 1.6 xx 10^(-6) s^(-1)`